You have found the following ages (in years) of all 5 gorillas at your local zoo: $ 14,\enspace 17,\enspace 13,\enspace 10,\enspace 15$ What is the average age of the gorillas at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 gorillas at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{14 + 17 + 13 + 10 + 15}{{5}} = {13.8\text{ years old}} $ Find the squared deviations from the mean for each gorilla. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $14$ years $0.2$ years $0.04$ years $^2$ $17$ years $3.2$ years $10.24$ years $^2$ $13$ years $-0.8$ years $0.64$ years $^2$ $10$ years $-3.8$ years $14.44$ years $^2$ $15$ years $1.2$ years $1.44$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.04} + {10.24} + {0.64} + {14.44} + {1.44}} {{5}} $ $ {\sigma^2} = \dfrac{{26.8}}{{5}} = {5.36\text{ years}^2} $ The average gorilla at the zoo is 13.8 years old. The population variance is 5.36 years $^2$.